Answer
$$\frac{{dy}}{{dx}} = \frac{{6{x^3} + 15{x^2} - 12x - 29}}{{{{\left( {3x - 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\left( {x + 3} \right)\left( {x + 1} \right)\left( {x + 2} \right)}}{{3x - 1}} \cr
& {\text{Simplify the numerator}} \cr
& y = \frac{{\left( {{x^2} + 4x + 3} \right)\left( {x + 2} \right)}}{{3x - 1}} \cr
& y = \frac{{{x^3} + 2{x^2} + 4{x^2} + 8x + 3x + 6}}{{3x - 1}} \cr
& y = \frac{{{x^3} + 6{x^2} + 11x + 6}}{{3x - 1}} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^3} + 6{x^2} + 11x + 6}}{{3x - 1}}} \right] \cr
& {\text{By the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {3x - 1} \right)\left( {3{x^2} + 12x + 11} \right) - \left( {{x^3} + 6{x^2} + 11x + 6} \right)\left( 3 \right)}}{{{{\left( {3x - 1} \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{9{x^3} + 36{x^2} + 33x - 3{x^2} - 12x - 11 - 3{x^3} - 18{x^2} - 33x - 18}}{{{{\left( {3x - 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{6{x^3} + 15{x^2} - 12x - 29}}{{{{\left( {3x - 1} \right)}^2}}} \cr} $$
